What is the integral of (e^{4x})/2 ?

The integral of (e^{4x})/2 is 1/8 e^{4x}+C

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integral of (e^{4x})/2

\int \frac{e ^{4 x}}{2} d x

\frac{1}{8} e^{4 x} + C

Solution steps

\int \frac{e ^{4 x}}{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int e^{4 x} d x

Apply u-substitution

: \int e^{u} \frac{1}{4} d u

= \frac{1}{2} \cdot \int e^{u} \frac{1}{4} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \frac{1}{4} \cdot \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= \frac{1}{2} \cdot \frac{1}{4} e^{u}

Substitute back

u = 4 x

= \frac{1}{2} \cdot \frac{1}{4} e^{4 x}

Simplify

\frac{1}{2} \cdot \frac{1}{4} e^{4 x} : \frac{1}{8} e^{4 x}

= \frac{1}{8} e^{4 x}

Add a constant to the solution

= \frac{1}{8} e^{4 x} + C

x \to 0 lim (x \cdot ln (x))

x \to 3 lim (4^{f (x)})

\int_{1}^{2 x} \frac{y}{x} d x

\frac{d y}{d x}, x y = 2 y + x

f (x) = - x^{3} + x^{2} + 4 x - 2